009A Sample Final A, Problem 2 - Revision history

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MathAdmin: Created page with "2. Find the derivatives of the following functions:
(a) $f(x)=\f..."$

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Created page with "2. Find the derivatives of the following functions: (a) <math style="vertical-align: -45%;">f(x)=\f..."

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2. Find the derivatives of the following functions:
 
   (a)  <math style="vertical-align: -45%;">f(x)=\frac{3x^{2}-5}{x^{3}-9}.</math>
 
   (b)  <math style="vertical-align: -15%;">g(x)=\pi+2\cos(\sqrt{x-2}).</math>
 
   (c)  <math style="vertical-align: -25%;">h(x)=4x\sin(x)+e(x^{2}+2)^{2}.</math>
 

{| class="mw-collapsible mw-collapsed" style = "text-align:left;
!Foundations:  
|-
|These are problems involving some more advanced rules of differentiation. In particular, they use
|-
|'''The Chain Rule:''' If ''f'' and ''g'' are differentiable functions, then
|-

|      <math>(f\circ g)'(x) = f'(g(x))\cdot g'(x).</math>
|-
| '''The Product Rule:''' If ''f'' and ''g'' are differentiable functions, then
|-
|      <math>(fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).</math>
|-
| '''The Quotient Rule:''' If ''f'' and ''g'' are differentiable functions and ''g''(''x'') ≠ 0, then
|-
|      <math>\left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. </math>
|-
|}

 '''Solution:'''

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Part (a):  
|-
|We need to use the quotient rule:
|-
|     <math>f'(x) = \frac {\left(3x^{2}-5\right)' \cdot \left(x^{3}-9 \right)- \left( 3x^{2}-5 \right) \cdot \left( x^{3}-9\right)'}{\left(x^{3}-9\right)^2} </math>
|-
|                 <math>= \frac{6x(x^{3}-9)-(3x^{2}-5)(3x^{2})}{(x^{3}-9)^{2}}</math>
|-
|                 <math>= \frac{6x^{4}-54x-9x^{4}+15x^{2}}{(x^{3}-9)^{2}}</math>
|-
|                 <math>= \frac{-3x^{4}+15x^{2}-54x}{(x^{3}-9)^{2}}.</math>
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Part (b):  
|-
|Both parts (b) and (c) attempt to confuse you by including the familiar constants ''e'' and π. Remember - they are just constants, like 10 or 1/2. With that in mind, we really just need to apply the chain rule to find
|-
|          <math>g'(x)=0-2\sin\left(\sqrt{x-2}\right)\cdot\frac{1}{2}\cdot(x-2)^{-1/2}\cdot1=\,-\frac{\sin\left(\sqrt{x-2}\right)}{\sqrt{x-2}}.</math>
|-
|
|}

{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
!Part (c):  
|-
|We can choose to expand the second term, finding
|-
|          <math>e(x^{2}+2)^{2}=ex^{4}+4ex^{2}+4e.</math>
|-
|We then only require the product rule on the first term, so
|-
|          <math>h'(x)=(4x)'\cdot\sin(x)+4x\cdot(\sin(x))'+\left(ex^{4}+4ex^{2}+4e\right)'=4\sin(x)+4x\cos(x)+4ex^{3}+8ex.</math>
|}

[[009A_Sample_Final_A|'''Return to Sample Exam''']]

@@ Line 22: / Line 22: @@
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).</math>
 |-
-|<br>'''The Quotient Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions and <math style="vertical-align: -22%;">g(x) \neq 0</math>, then
+|<br>'''The Quotient Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions and <math style="vertical-align: -21%;">g(x) \neq 0</math>&thinsp;, then
 |-
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. </math>

@@ Line 22: / Line 22: @@
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).</math>
 |-
-|<br>'''The Quotient Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions and <math style="vertical-align: -25%;">g(x) \neq 0</math>, then
+|<br>'''The Quotient Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions and <math style="vertical-align: -22%;">g(x) \neq 0</math>, then
 |-
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. </math>
@@ Line 50: / Line 50: @@
 |Both parts (b) and (c) attempt to confuse you by including the familiar constants <math style="vertical-align: 0%;">e</math> and <math style="vertical-align:  0%;">\pi</math>. Remember - they are just constants, like 10 or 1/2.  With that in mind, we really just need to apply the chain rule to find
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>g'(x)=0-2\sin\left(\sqrt{x-2}\right)\cdot\frac{1}{2}\cdot(x-2)^{-1/2}\cdot1=\,-\frac{\sin\left(\sqrt{x-2}\right)}{\sqrt{x-2}}.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>g'(x)\,=\,0-2\sin\left(\sqrt{x-2}\right)\cdot\frac{1}{2}\cdot(x-2)^{-1/2}\cdot1\,=\,\,-\,\frac{\sin\left(\sqrt{x-2}\right)}{\sqrt{x-2}}.</math>
 |-
+|
@@ Line 64: / Line 64: @@
 |We then only require the product rule on the first term, so
 |-
-|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>h'(x)=(4x)'\cdot\sin(x)+4x\cdot(\sin(x))'+\left(ex^{4}+4ex^{2}+4e\right)'=4\sin(x)+4x\cos(x)+4ex^{3}+8ex.</math>
+|&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; <math>h'(x)\,=\,(4x)'\cdot\sin(x)+4x\cdot(\sin(x))'+\left(ex^{4}+4ex^{2}+4e\right)'\,=\,4\sin(x)+4x\cos(x)+4ex^{3}+8ex.</math>
 |}
 [[009A_Sample_Final_A|'''<u>Return to Sample Exam</u>''']]

@@ Line 26: / Line 26: @@
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. </math>
 |-
 |}

@@ Line 13: / Line 13: @@
 |These are problems involving some more advanced rules of differentiation.  In particular, they use
 |-
-|'''The Chain Rule:''' If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -20%;">g</math> are differentiable functions, then
+|'''The Chain Rule:''' If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
 |-
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(f\circ g)'(x) = f'(g(x))\cdot g'(x).</math>
 |-
-|<br>'''The Product Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -20%;">g</math> are differentiable functions, then
+|<br>'''The Product Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions, then
 |-
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>(fg)'(x) = f'(x)\cdot g(x)+f(x)\cdot g'(x).</math>
 |-
-|<br>'''The Quotient Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -20%;">g</math> are differentiable functions and <math style="vertical-align: -25%;">g(x) \neq 0</math>, then
+|<br>'''The Quotient Rule:'''  If <math style="vertical-align: -25%;">f</math> and <math style="vertical-align: -15%;">g</math> are differentiable functions and <math style="vertical-align: -25%;">g(x) \neq 0</math>, then
 |-
 |<br>&nbsp;&nbsp;&nbsp;&nbsp; <math>\left(\frac{f}{g}\right)'(x) = \frac{f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^2}. </math>

@@ Line 8: / Line 8: @@
 <br>
-{| class="mw-collapsible mw-collapsed" style = "text-align:left;
+{| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Foundations: &nbsp;
 |-